Just Divide and Conquer.

Relevant wiki: Sum of n, n², or n³

Let $\displaystyle S_n = \sum_{k=1}^n \frac {\lambda k^4+2k^3+k^2+k+1}{3n^5+n^2+n+5k}$ .

As $n \to \infty$ ,

$\begin{aligned} S_{n \to \infty} & \to \frac {\sum_{k=1}^n \lambda k^4}{3n^5} = \frac {\lambda n(n+1)(2n+1)(3n^2+3n-1)}{30\times 3n^5} \to \frac {6\lambda n^5}{90n^5} = \frac \lambda {15} = \frac 13 \\ \implies \lambda & = \boxed{5} \end{aligned}$

Let $x_k = k/n$ . Then, dividing the numerator and the denominator by $n^4$ , we have $\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{n} \frac{\lambda x_k^4 + 2x_k^3 n^{-1} + x_k^2 n^{-2} + x_k n^{-3} + n^{-4}}{3 + n^{-2} + n^{-3} + 5 x_k n^{-3}} = \lim_{n\to\infty} \sum_{k=1}^n \frac{1}{n} \frac{\lambda x_k^4}{3}$

In the last equality, we observed that each term of the numerator that had a negative power of $n$ was zero when taken the limit.

The last expression is the Riemman sum of the function $f(x) = \lambda x^4 / 3$ , which converges to the integral from $0$ to $1$ of the function $f$ . Thus, $\begin{aligned} \int_0^1 \frac{\lambda x^4}{3} dx & = \frac{1}{3}, \\ \frac{\lambda x^5}{5} \Big|_0^1 & = 1 \\ \lambda & = 5. \end{aligned}$